Consider the Fibonacci sequence in mod :
. Now,
consider the pairs of consecutive Fibonacci numbers (for
example , , etc.). Since we are in mod , there are only
possible distinct such pairs (since there are choices for the first
number in a pair and choices for the second number in a pair). If is
one such pair, then the whole sequence must be
in mod ,
which contradicts the initial conditions of the sequence, so there are only
possible distinct pairs. Since the Fibonacci sequence is infinite,
there are an infinite number of pairs, so by the Pigeonhole Principle, at least
one of these pairs must repeat. Suppose that the pairs
and
are the same (that is,
and
). WLOG (without loss of generality), let
. Then, we know that
and
, so
.
We can continue this until we find that
and
, so . So, we can continue
this in the other direction and find that the Fibonacci sequence is periodic.

By the definition of the Fibonacci sequence, we know that
. So,
.
Then,
. Also,
. Notice that the coefficients of and
are being generated recursively with the same formula as the Fibonacci numbers,
so we can say that
if . So,
we can substitute for to say that
. Now, since we know that
(by hypothesis), then for some . So, we
can substitute to get
. Now, we can
factor out an to say
. So, by
definition of divides, , and since , then
, which is what we wanted to show.

For , we want to show that
.
Since we know that
and (by definition), so
, and
. For ,
we want to show that
. By the definition of the
Fibonacci sequence,
, so
. Since
, then
. Since , we can multiply to get that
, which is what we wanted to show.

Induction Step: Assume
,
such that .

WTS:
.

Since and , we can say that
and
. We can add these equations together
to say that
.
Factoring, we get that
. Since
and
, we can substitute to say that
, which is what we wanted to show.

Let . Divide by (by Fact 4) to get , where
. So,
. Since
,
then
. Since
(by definition
of , then , and since (by Theorem 2),
then , so
. So we can apply Theorem
3 to say that
. Since
, so
. By Theorem
4,
. Since , then by Fact
7,
. Since , then by definition of
, it follows that
. So, by Fact 7,
. Now, since
, then by Fact
14,
. So, if , then satisfies the
definition of and is less than , which is a contradiction. So,
must be equal to , so , so , so .

Let and and . Since , then
and , where
(by definition of LCM). So,
and . Also,
and
by
definition of and , so and . So, by Theorem
2, and . Since , by Fact 8,
, so
, so
,
which is what we wanted to show.

Part 2: WTS: If
, then
.

Let and and . Since , then
and , where
(by definition of LCM). Since
, then , so and . So,
and
. Now, by Theorem
5, and . So, by definition of LCM, , so
(by Theorem 3). So,
, which is what we wanted to show, so we are done.

Let . Divide by to get , where and . Since
, then
, so
(by Theorem 8). Since
, then
, so
(by
Theorem 8). If , then satisfies the definition of
, but is smaller than (since ), which is a contradiction.
So, , and , so .

We will show this by showing that
and
.
and
(by definition of LCM), and since
, then
(by Theorem 10).
and
, so
and
by Fact
16, so by Theorem 9, . Since
and
, so
and
by Fact
16, so by Theorem 9, . So, by
definition of LCM,
. Since and ,
by Fact 15,
.

Let , so (by definition). By Theorem
4,
, so for some , so by
Theorem 7, . Therefore, exists, so
exists. Now, if we find that the first directly before a is in
position number , then , since
and
. So, we must examine position
numbers
, since all zeroes are in position
numbers
(by Theorem 5). Now, by Theorem
7,
(since
by definition of ). So, we are looking for
the first such that
. By definition,
this is
. So,
, or
.